Prove or disprove that $f(x)\leq 0$

Question

Problem :

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Let $x,a>0$ then (dis)prove :

$$f(x)=\left(1+x\right)\ln\left(x\right)-\left(\left(x+a\right)\ln\left(\frac{x+a}{2}\right)+\left(x-a\right)\left(1+\frac{\frac{1}{12}\left(x-a\right)^{2}}{\left(x+a\right)^{2}}\right)\right)\leq 0$$

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My attempt :

Using this question Prove that $\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$ we have :

$$g(x)=2^{\frac{3}{4}}\left(x\right)^{\frac{1}{4}}\left(x^2+1\right)^{\frac{1}{4}}-\left(x+1\right)\leq 0 $$

Then and without success I tried to show $0<x\leq 10$:

$$f(x)\leq g(x)$$

Perhaps we can split the problem.

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Question :

How to prove or disprove it ?

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Reference :

Prove $2(x + y)\ln \frac{x + y}{2} - (x + 1)\ln x - (y + 1)\ln y \ge 0$

Harnack's inequality in the plane R^2

See also Boltzmann's entropy law

Edit 19/02/2022 :

For $a\geq 1\geq x >0$ we have the inequality :

$$r\left(x\right)=b\left(\frac{b}{2}+\frac{1}{cb\left(2\sqrt{x}-b+2\right)^{2}}\right)\cdot2\left(x-1\right)+\frac{\left(1-x\right)^{2}}{6}-\frac{\left(1-x\right)^{3}}{32}\geq \left(x+1\right)\ln\left(x\right)$$

Where : $b=\frac{4}{3},c=-\frac{2}{\left(b-4\right)^{2}\left(b^{2}-2\right)}$

And :

$$r(x)\leq \left(x+a\right)\ln\left(\frac{\left(x+a\right)}{2}\right)+\left(x-a\right)\left(1+\frac{1}{12}\frac{\left(x-a\right)^{2}}{\left(x+a\right)^{2}}\right)$$

Let :

$d(x,y)=\left(x+y\right)\ln\left(\frac{\left(x+y\right)}{2}\right)+\left(x-y\right)\left(1+\frac{1}{12}\frac{\left(x-y\right)^{2}}{\left(x+y\right)^{2}}\right)-r(x)$

Then define :

$$t(x,y)=\frac{d}{dx}\left(\frac{d\left(x^{2},y\right)}{x^{2}+y}\right)$$

Then :

$$t(1/(x+1),y+1)=(-3072x^{14}y^{3}-9216x^{14}y^{2}-9216x^{14}y-3072x^{14}-43979x^{13}y^{3}-133946x^{13}y^{2}-135955x^{13}y-45988x^{13}-291698x^{12}y^{3}-922364x^{12}y^{2}-969634x^{12}y-338968x^{12}-1186833x^{11}y^{3}-3999862x^{11}y^{2}-4443243x^{11}y-1630214x^{11}-3304964x^{10}y^{3}-12227800x^{10}y^{2}-14617996x^{10}y-5695160x^{10}-6653705x^{9}y^{3}-27893214x^{9}y^{2}-36430788x^{9}y-15193288x^{9}-9974034x^{8}y^{3}-48917820x^{8}y^{2}-70589208x^{8}y-31678512x^{8}-11300367x^{7}y^{3}-66863754x^{7}y^{2}-107366388x^{7}y-52021704x^{7}-9729768x^{6}y^{3}-71392560x^{6}y^{2}-127926432x^{6}y-67057728x^{6}-6358044x^{5}y^{3}-59115000x^{5}y^{2}-117970224x^{5}y-66977184x^{5}-3132848x^{4}y^{3}-37309952x^{4}y^{2}-82324480x^{4}y-50639488x^{4}-1150160x^{3}y^{3}-17402480x^{3}y^{2}-41927296x^{3}y-27916288x^{3}-306624x^{2}y^{3}-5679424x^{2}y^{2}-14685696x^{2}y-10551296x^{2}-55040xy^{3}-1164544xy^{2}-3168256xy-2441216x-5120y^{3}-113664y^{2}-319488y-262144)/(144(x+1)^{3}(x+4)^{3}(x^{2}y+x^{2}+2xy+2x+y+2)^{4})\leq 0$$

For the case $x\geq 1\ge a >0$ one can show that :

$$v\left(x\right)=b\left(\frac{b}{2}+\frac{1}{cb\left(2\sqrt{x}-b+2\right)^{2}}\right)\cdot2\left(x-1\right)+\frac{\left(1-x\right)^{2}}{6}+\frac{\frac{1}{4}\left(x-1\right)^{3}}{\left(x+1\right)^{3}}-\left(x+1\right)\ln\left(x\right)\geq 0$$

Now considering :

$$m(x)=f\left(x\right)-v\left(x\right)\geq 0$$

Then define :

$$T(x,a)=\frac{d}{dx}\left(\frac{m\left(x^{2}\right)}{x^{2}+a}\right)$$

Then $x>1>y>0$ $$T(1+x/(x+y),1-y/(x+y))=((20394250x^{20}+528955025x^{19}y+4207383600x^{18}y^{2}+17983502665x^{17}y^{3}+49580654654x^{16}y^{4}+95957785654x^{15}y^{5}+136004398310x^{14}y^{6}+143361431309x^{13}y^{7}+110895037540x^{12}y^{8}+58405263383x^{11}y^{9}+14075272220x^{10}y^{10}-7938105344x^{9}y^{11}-11533507934x^{8}y^{12}-7506291198x^{7}y^{13}-3305576312x^{6}y^{14}-1061515832x^{5}y^{15}-251347968x^{4}y^{16}-43018176x^{3}y^{17}-5054208x^{2}y^{18}-365568xy^{19}-12288y^{20})/(6(x+y)(7x+4y)^{3}(5x^{2}+5xy+y^{2})^{4}(5x^{2}+6xy+2y^{2})^{4}))>0$$

Answer 1

Another strategy :

If :

$$f\left(x\right)=\left(1+x\right)\ln\left(x\right)-\left(\left(x+a\right)\ln\left(\frac{x+a}{2}\right)+\left(x-a\right)\left(1+\frac{\frac{1}{12}\left(x-a\right)^{2}}{\left(x+a\right)^{2}}\right)\right)$$

Then for $x\geq 1,a\in(0,1]$ :

$$f''(x)\leq 0 $$

Proof :

we have the inequalities :

$$\frac{4a^{3}}{(a+x)^{4}}-\frac{2a^{2}}{(a+x)^{3}}-\left(1-a\right)\frac{1}{x^{2}}<0$$

And :

$$\frac{-1}{a+x}-ax^{-2}+\frac{1}{x}<0$$

The sum of the two inequalities is the second derivative .

And then for $1/4\leq a\leq 1$ :

$$f'(2-a+\frac{1}{2}\left(a-1\right)^{2})\leq 0$$

Proof :

we have :

$$h\left(x\right)=\log((2x)/(a+x))+1/x,f\left(x\right)=-\frac{\left(a^{2}(a-3x)\right)}{(3(a+x)^{3})}-\frac{1}{12}$$

And we have :

$$f\left(2-a+\frac{1}{2}\left(a-1\right)^{2}\right)\leq 0,h\left(\left(2-a+\frac{1}{2}\left(a-1\right)^{2}\right)\right)-1\leq 0$$

Finally we have :

$$f'(y)=0$$ then we have $1/4\leq a\leq 1$:

$$y\in[2-a,2-a+\frac{1}{2}\left(a-1\right)^{2}]$$

To conclude partially we have $1/4\leq a\leq 1$:

$$f'\left(2-a+\frac{1}{2}\left(a-1\right)^{2}\right)\left(x-\left(\left(2-a\right)+\frac{1}{2}\left(a-1\right)^{2}\right)\right)+f\left(\left(2-a\right)+\frac{1}{2}\left(a-1\right)^{2}\right)\leq 0$$ because for $0<y\leq 1$:

$$((-6\ln(((y^{2}-4y+5)/(2)))y^{9}+78\ln(((y^{2}-4y+5)/(2)))y^{8}-516\ln(((y^{2}-4y+5)/(2)))y^{7}+2244\ln(((y^{2}-4y+5)/(2)))y^{6}-6960\ln(((y^{2}-4y+5)/(2)))y^{5}+15936\ln(((y^{2}-4y+5)/(2)))y^{4}-26940\ln(((y^{2}-4y+5)/(2)))y^{3}+32700\ln(((y^{2}-4y+5)/(2)))y^{2}-26250\ln(((y^{2}-4y+5)/(2)))y+11250\ln(((y^{2}-4y+5)/(2)))+15y^{9}-12y^{8}\ln(((y^{2}-2y+5)/(4)))-193y^{8}+120y^{7}\ln(((y^{2}-2y+5)/(4)))+1258y^{7}-672y^{6}\ln(((y^{2}-2y+5)/(4)))-5138y^{6}+2472y^{5}\ln(((y^{2}-2y+5)/(4)))+14328y^{5}-6504y^{4}\ln(((y^{2}-2y+5)/(4)))-28548y^{4}+12360y^{3}\ln(((y^{2}-2y+5)/(4)))+41238y^{3}-16800y^{2}\ln(((y^{2}-2y+5)/(4)))-42110y^{2}+15000y\ln(((y^{2}-2y+5)/(4)))+28025y-7500\ln(((y^{2}-2y+5)/(4)))-8875)/(6(y^{2}-4y+5)(y^{2}-2y+5)^{3})))\leq 0$$

$$-f'\left(x+y\right)y+f\left(x+y\right)\leq 0$$

For $x \in[2-a,2-a+\frac{1}{2}\left(a-1\right)^{2}]$

So the maximum is negative or zero .

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We have (see the reference) for $a\geq x\geq 1$ :

$$12\ln\left(\frac{e}{2}\right)\left(x+1\right)\ln\left(x\right)\le 12\ln\left(\frac{e}{2}\right)\left(x+a\right)\ln\left(\frac{x+a}{2}\right)+\left(x-a\right)\left(12\ln\left(\frac{e}{2}\right)+\frac{\left(x\ln x-a\ln a\right)}{x-a}-1\right)\leq 12\ln\left(\frac{e}{2}\right)\left(\left(x+a\right)\ln\left(\frac{x+a}{2}\right)+\left(x-a\right)\left(1+\frac{\frac{1}{12}\left(x-a\right)^{2}}{\left(x+a\right)^{2}}\right)\right)$$

Reference :

@article{Kouba2018SharpTB, title={Sharp two-parameter bounds for the identric mean}, author={Omran Kouba}, journal={Journal of Inequalities and Applications}, year={2018}, volume={2018} }

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We have for $x\geq 3$ and $0\leq a\leq 0.1875$ :

$$\left(x+a\right)\ln\left(\frac{x+a}{2}\right)+\left(x-a\right)\left(1+\frac{1}{12}\frac{\left(x-a\right)^{2}}{\left(x+a\right)^{2}}\right)> g\left(x-c\right)>\left(x+1\right)\ln\left(x\right)$$

Where $$c=a-\frac{8}{100}+\frac{1}{2}\left(a-\frac{1}{4}\right)^{2},g\left(x\right)=\left(x+a\right)\ln\left(x+\frac{\left(1+a\right)2}{5}x\right)$$

Answer 2

Alternative formulation and partial results

Proving $f(x)\le 0$ is equivalent to proving $f(ax)\le 0$, since $$ \{(x,a):x>0\ \ , \ \ a>0\}=\{(ax,a):x>0\ \ , \ \ a>0\}. $$ For this reason, we have $$ f(ax){=\left(1+ax\right)\ln\left(ax\right)-\left(a(x+1)\ln\left(\frac{ax+a}{2}\right)+a\left(x-1\right)\left(1+\frac{\frac{1}{12}\left(x-1\right)^{2}}{\left(x+1\right)^{2}}\right)\right) \\= \ln\left(ax\right)+ax\ln\left(\frac{2x}{x+1}\right)-\left(a\ln\left(\frac{ax+a}{2}\right)+a\left(x-1\right)\left(1+\frac{\frac{1}{12}\left(x-1\right)^{2}}{\left(x+1\right)^{2}}\right)\right) \\= (1-a)\ln\left(ax\right)+a(1+x)\ln\left(\frac{2x}{x+1}\right)-\left(a\left(x-1\right)\left(1+\frac{\frac{1}{12}\left(x-1\right)^{2}}{\left(x+1\right)^{2}}\right)\right). } $$ Therefore, we need to (dis)prove that $$ (1+x)\ln\left(\frac{2x}{x+1}\right)-\left(x-1\right)\left(1+\frac{\frac{1}{12}\left(x-1\right)^{2}}{\left(x+1\right)^{2}}\right)\le \frac{a-1}{a}\ln\left(ax\right). $$ Also by a change of variables $u=\frac{x-1}{x+1}$ with $u\in (-1,1)$, an equivalent form of the inequality is $$ \ln\left(1+u\right)-u-\frac{u^3}{12}\le \frac{(a-1)(1-u)}{2a}\ln\left(a\frac{1+u}{1-u}\right). $$

Partial result

It is easy to prove that $\ln\left(1+u\right)-u-\frac{u^3}{12}$ is a concave function with a maximum of $0$ at $u=0$. The inequality then holds for values of $u$ and $a$ for which $\frac{(a-1)(1-u)}{2a}\ln\left(a\frac{1+u}{1-u}\right)\ge 0$, i.e. when: $$ \begin{cases} a-1\ge 0\quad,\quad \ln\left(a\frac{1+u}{1-u}\right)\ge 0 \implies a\ge 1\quad,\quad ax\ge 1 \\ a-1\le 0\quad,\quad \ln\left(a\frac{1+u}{1-u}\right)\le 0 \implies a\le 1\quad,\quad ax\le 1. \end{cases} $$

Update

Using the Taylor's series, it is possible to proceed into (perhaps) more simplified version of the inequality. Sparing the results, the final series is $$ \frac{(4-a)u^3}{12} + \sum_{n=2,even}^\infty \frac{a+n(1-2a)}{n(n-1)}u^n + \sum_{n=5,odd}^\infty \frac{u^n}{n} \le \frac{a-1}{2} \left[(1-u)\ln a+2u \right]. $$ I wasn't been able so far to use it for solving the problem.